`1*2*3+2*3*4+...+n(n+1)(n+2)=(n(n+1)(n+2)(n+3))/4 forall n in N.`

Explanation

Let `P(n):(1)/(1.2.3)+(1)/(2.3.4)+.....+(1)/(n(n+1)(n+2))=(n(n+3))/(4(n+1)(n+2))` .....(i)
Step I For n=1.
LHS is Eq. (i) `=(1)/(1.2.3)=(1)/(6)` and RHS of Eq. (i) `=(1(1+3))/(4(1+1)(1+2))=(1)/(6)`
Therefore , P(1) is true .
Step II Assume that P(k) is ture , then `P(k):(1)/(1.2.3)+(1)/(2.3.4)+......+(1)/(k(k+1)(k+2))=(k(k+3))/(4(k+1)(k+2))`
Step III For `n=k+1`,
``P(k):(1)/(1.2.3)+(1)/(2.3.4)+.....+(1)/(k(k+1)(k+2))+(1)/((k+1)(k+2(k+3)))=((k+1)(k+4))/(4(k+2)(k+3))`
`therefore LHS =(1)/(1.2.3)+(1)/(2.3.4)+.....+(1)/(k(k+1)(k+2))+(1)/((k+1)(k+2)(k+3))`
` =(k(k+3))/(4(k+1)(k+2))+(1)/((k+1)(k+2)(k+3))` [by assumption step ]
`=(k(k+3)^2+4)/(4(k+1)(k+2)(k+3))`
`=(k^3+6k^2+9k+4)/(4(k+1)(k+2)(k+3))`
`=((k+10^2(k+4))/(4(k+1)(k+2)(k+3))`
`=((k+1)(k+4))/(4(k+2)(k+3))=RHS`
Therefore , `P(k+1)` is true , Hence , by the principle of mathematical P(n) is true for all `n in N`.